INDEX

Chapter-1 Real Numbers

Introduction

Since our childhood we have been using 4 fundamental operation addition , subtraction , division and multiplication and we have applied these number on different type of numbers till now  and In Class 9 we have studied about real numbers specially about irrational number. in this section we will learn chapter-1:Real Numbers of Class 10. and we will begin with Euclid’s Division lemma and fundamental theorem of arithmetic.
Euclid’s division lemma tells us about the divisibility(it means how does any integers divides the other integers and tells us about the remainder and quotient) of integers. it is quit easy to state and  understand it. it state that any positive integers a can be divided by any other positive integer in such a way that it leaves remainder r which is smaller than b. Euclid’s division lemma helps us in finding the HCF of two positive integers.
Fundamental theorem of arithmetic tells about expressing positive integers as the product of prime integers.it states that every positive integers is prime number or it can be factorized as the product of power of prime numbers. in this chapter we will apply this theorem to prove the irrationality of numbers like \sqrt2 , \sqrt3 , \sqrt5 etc . we know that decimal representation of rational number is either terminating or if it is non-terminating then it is repeating. The prime factorization of the denominator of rational number completely reveals the nature of its decimal representation. here we will discuss in details but if you need to get shorts notes of real number then you can visit here.

Now lets began with divisibility of integers :-

Divisibility

          DEFINITION :-

A non-zero integers ‘a’ is said to be divide an integers ‘b’ if there exist an integer ‘c’ such that b = ac . The integer ‘b’ is called dividend , integer ‘a’ is known as the divisor and integer ‘c’ is known as the quotient.

for example :- 3 divides 36 because there is an integer 12 such that 36 = 3 \times 12 however 3 does not divides 35 because there do not exist an integer ‘c’ such that 35 = 3 \times c . it means 35 = 3 \times c is not true for any integer c

if a non zero integer ‘a’ divides an integer b then we can write this as a \mid b (this is read as “a divides b”) then we can  say that

  • b is divisible by a or
  • a is factor of b or
  • b is a multiple of a or
  • a is a divisor of b

if it is written like a \nmid b it means b is not divisible by a
(i) -4 \mid 20 , because there exists an integer – 5 such that 20 = -4 \times (-5) .
(ii) 4 \mid -20 , because there exists an integer – 5 such that – 20 = 4 \times (-5)
(iii) -4 \mid -20 , because there exists an integer  5 such that – 20 = -4 \times 5

Example :-  which of the following is true or not true.
(a) 3 \mid 93
(b) 6 \mid 28
(c) 0 \mid 4

Solution :- (a)  3 \mid 93 is true because 93 = 3 \times 31
(b) 6 \mid 28 is not true , because 28 = 6c is not valid for any integers
(c) 0 \mid 4 is not true by definition because it will give not define solution(infinity).

Properties of Divisibility :- 

  • \pm 1 divides every non-zero integer.

i.e :- \pm 1 \mid a for every non – zero integer a

  • 0 is divisible by every non – zero integer a .

i.e :-  a \mid 0 for every non – zero integer a.

  • 0 does not divide any integer.
  • if a is a  non-zero integer and b is any integer ,

then a \mid b  \Rightarrow  a \mid -b , -a \mid b   and -a \mid -b

  • if a and b are non-zero integers , then  a \mid b   and   b \mid a  \Rightarrow , a =  \pm b   
  • if a and c are non-zero integer and b, d are any two integers then,

(i) a \mid b  and  c \mid d \Rightarrow ac \mid bd .
(ii) ac \mid bc \Rightarrow a \mid b  

 

Euclid's division lemma

Euclid was the first greek mathematician who initiated a new way of thinking the study of geometry. he made the important contribution in the number theory one of them is “Euclid’s lemma” .

LEMMA :- it is proven statement which is used to prove other statements.

consider the division of one positive integer to another positive example :- 58 by 9 . in this case we can write that 58 = 9 \times 6 + 4 , where  0 \le 4 < 9

So , For any two given two positive integers a and b , there exits unique whole number q and r such that a = bq + r , where 0 \le r < b

we can also call it as

  • a = dividend
  • b = divisor
  • q = quotient
  • r = remainder

Dividend = (divisor \times quotient) + remainder.

Proof of Euclid's division lemma

proof :-

Consider the following arithmetic progression

……. (a – 3b) , (a – 2b) , (a – b) , a , (a + b) , (a + 2b) , (a + 3b) , ………

Clearly , it is an arithmetic progression with common difference ‘b’ and it extends indefinitely in both directions. let r be the  smallest non-negative term of this arithmetic progression. then there exists a non-negative integer q such that a – bq = r \Rightarrow a = bq + r. as r is the smallest non-negative integer. therefore 0 \le r < b

we have ,a = bq + r , where  0 \le r < b , we will check the uniqueness of q and r .

uniqueness :- to prove the uniqueness of q and r , let us assume that there is another pair q_1 and r_1 of non-negative integer satisfying the same relation. a = bq_1 + r_1 , where   0 \le r_1 < b ,  we need to proove r_1 = r  and  q_1 = q then ,

a = bq + r  and a = bq_1 + r_1

\Rightarrow  bq + r = bq_1 + r_1

\Rightarrow  r_1 – r = bq – bq_1

\Rightarrow  r_1 – r = b(q – q_1)

\Rightarrow  b\mid r_1 – r

\Rightarrow  r_1 – r = 0       [ \because 0 \le r < b  and  0 \le r_1 < b  \Rightarrow 0 \le r_1 – r < b]

\Rightarrow  r_1 =  r

now ,    r_1 =  r

\Rightarrow  -r_1 =  – r                   [multiplying both side by (-1)]

\Rightarrow a –  r_1 = a –  r                                       [adding a both sides]

  bq_1 =  bq                                      \binom {\because a = bq + r  and  a = bq_1 + r_1}{\because bq = a – r  and  bq_1 = a – r_1}

\Rightarrow  bq_1 =  bq

hence , the representation a = bq + r , 0 \le r < b is unique.

REMARK -1 :- the above lemma is nothing but a restatement of the long division process we have been doing all these years and that the integers q and r are called quotient and remainder respectively.

REMARK – 2 :- the above lemma has been stated for positive integer only , but it can be extended to all integers.

REMARK – 3 :- (i) when a positive integer is divided by 2 , the remainder is either 0 or 1 . so any positive integer is of the form 2m , 2m+1 for some integer m.
(ii) when any positive integer is divided by 3 , the remainder is 0 or 1 or 2. so any positive integer can be written in the form of 3m , 3m+1 , 3m+2 .

Example -1 :- show that any positive odd integer is of the form 4q + 1  or 4q + 3 , where q is some integer.

Solution :- let a be any odd positive integer and b = 4 , by division Euclid’s division lemma there exists a integers q and r such that a = 4q + r  , where  0 \le r < 4  

\Rightarrow a = 4q  or,  a = 4q + 1  or,  a = 4q + 2  or,  a = 4q + 3 [\because 0 \le r < 4 \Rightarrow r = 0,1,2,3]

\Rightarrow a = 4q + 1  or,  a = 4q = 3           [\because  a is an odd integer \therefore a \ne 4q , a \ne 4q + 2 ]

hence any odd integer is of the form 4q + 1 or , 4q + 3 .

Algorithm

An algorithm is a series of well defined steps which gives a method for solving a certain type of problem and process of calculation repeated successively on the result of earlier steps till the desired result obtained.

Euclid's division algorithm :-

It is a technique to compute the HCF of two given positive integers , say a and b with a > b , in the following steps.:-

  1. on dividing a by b , we get the quotient q and remainder r such that a = bq + r , where , 0 \le r < b .
  2. if r = 0 then HCF(a , b) = b  , if r \ne 0 then apply the division lemma to b and r 
  3. continue the process till the remainder is 0 . the last divisor will be the required HCF.

Or,

THEOREM :- if a and b are positive integer such that a = bq + r , every common divisor of a and b is a common of b and r and vice – versa 

Proof of Euclid's division lemma
let c be a common divisor of a and b the ,
c \mid a \Rightarrow a = cq_1         for  some  integer   q_1           c \mid b \Rightarrow  b  = cq_2  for   some  integer  q_2
Now ,
a = bq + r
\Rightarrow   r = a -bq
\Rightarrow   r = cq_1 -cq_2 q
\Rightarrow   r = c(q_1 – q_2 q)
\Rightarrow   c \mid r
\Rightarrow   c \mid r  and   c \mid b
c is common divisor of b and r
hence . a common divisor of a and b is a common divisor of b and r.
Conversely ,
let d be a common divisor of b and r then,
d \mid b \Rightarrow b = r_1 d  for  some  integer  r_1
d \mid r  \Rightarrow  r = r_2 d  for  some  integer  r_2
we will now show that d is a common divisor of a and b .
we have ,
a = bq + r
\Rightarrow  a = r_1 d q + r_2 d \Rightarrow  a = (r_1  q + r_2) d \Rightarrow  d \mid a \Rightarrow  d \mid a     and    d \mid b  \Rightarrow   d is common divisor of a and b. [proved]

Example :- use Euclid’s algorithm to find the HCF of 272 and 1032.

Solution :- we find HCF (272 , 1032 ) using the following steps :- 

Step 1 :- since 1032 > 272 , we divide 1032 by 272 to get 3 as quotient and 216 as remainder. so by Euclid’s division lemma we get 1032 = 272 \times 3 + 216.

Step 2 :- since the remainder 216 \ne 0 , we divide 272 by 216 to get 1 as quotient and 56 as remainder.          \therefore by Euclid’s  division lemma , we get 272 = 216 \times 1 + 56

step 3 :- since the remainder 56 \ne 0 , we divide 216 by 56 to get 3 as quotient and 48 as remainder.          \therefore by Euclid’s division lemma , we get 216 = 56 \times 3 + 48 .

step 4 :- since the remainder 48 \ne 0 , we divide 56 by 48 to get 1 as quotient and 8 as remainder.            \therefore by Euclid’s division lemma , we get 56 = 48 \times 1 + 8

step 5 :- since the remainder 8 \ne 0 , we divide 48 by 8 to get 6 as quotient and 0 as remainder.            \therefore by Euclid’s division lemma , we get 48 = 8 \times 6 + 0.

the remainder has now become 0, so our procedure stops.

hence , HCF(272 , 1032) = 8

Chapter-1 Real Numbers

REMARK :- 

The HCF of numbers is a common divisor of the numbers which are divisor of their LCM , Consequently , HCF is a divisor of LCM.

Step 1 :- Apply Euclid’s division lemma to a and b and obtain whole number q_1   and r_1 such that a = b q_1  +  r_1  , 0 \le  r_1 < b

step 2 :- if  r_1   = 0 , b is the HCF of a and b 

step 3 :- if r_1 \ne 0 , apply Euclid’s division lemma to b and r_1 and obtain two whole numbers q_1  and  r_2    such that b = q_1 r_1  +  r_2

step 4 :- if r_2 = 0 then r_1 is the HCF of a and b .

step 5 :- if r_2  \ne 0  , then apply Euclid’s division lemma to r_1 and r_2 and continue the above process. till the  r_n is zero. the divisor at this stage i.e. r_n-1 or the non-zero remainder at the previous stage is the HCF of a and b

Rule for Finding HCF of three numbers

Step 1 :- Find the HCF of any two of the given numbers.

Step 2 :- Find the HCF of any third given number and the HCF obtained in step 1.

Step 3 :- the HCF obtained in the step 2 is the HCF of three given numbers.

Fundamental Theorem of Arithmetic

Theorem – 1:- Every composite number can be expressed as a product of primes , and this factorization is unique except for the order i which the prime factor occur.

Theorem – 2 :- Let p be a prime number and a be a positive integer. if p divides a^2 , then p divides a. 

Proof :- 

from the fundamental theorem of Arithmetic integer a can be factorized as the product of primes. let a = p_1 p_2 p_3 …… p_n be the prime factorization of a where , p_1  ,  p_2  ,  …….  , p_n are prime , not necessarily distinct.

now , 

a = p_1 p_2 p_3 …… p_n

  \Rightarrow  a^2  =  (p_1 p_2 p_3 …… p_n) (p_1 p_2 p_3 …… p_n)   \Rightarrow  a^2  =  p_1^2 p_2^2 p_3^2 …… p_n^2

it is given that p is prime and it divides a^2 . therefore , p is a prime factor of a^2 . from the uniqueness part of the fundamental theorem of arithmetic it follows that the only prime factors of a^2 are p_1  ,   p_2  ,  p_3  , …… ,  p_n therefore p is one of   p_1  ,   p_2  ,  p_3  , …… ,  p_n . this implies that 

\Rightarrow p \mid p_1 p_2 p_3 …… p_n

\Rightarrow p \mid a .

Some Application of the Fundamental Theorem of Arithmetic

in this section we will see the use of fundamental theorem of arithmetic in expressing the given integers as the product of primes.

Finding HCF and LCM of positive integers.

following is the algorithm of finding the HCF and LCM of two or more positive integers :-

Algorithm :- 

Step – 1 :- factorize each of the given positive and express them as a product of powers of primes in ascending order of magnitudes of primes.

step-2 :- to find HCF , identify common prime factor and find the smallest(least) exponent of these common factors . now raise these common prime factors to their smallest exponents and multiply them to get the HCF .

    To find the LCM , list all the prime factors(once only) occurring in the prime factorization of the given positive integrs.

    for each  of these factor , find the greatest exponent and raise each prime factor to the greatest exponent and multiply them to get the LCM.

REMARK :- To find the LCM of two positive integers a and b , we can also use the following result , if we heave already found the HCF .

HCF \times LCM = a \times   b .

Irrational Numbers

The Number which when expressed in decimal form are expressible as non-terminating and non-repeating decimals are known as irrational number 

example :- 

Type 1 :- 

    • 0.1010010001……

Type  2 :- 

    • \sqrt{2}  ,  \sqrt{3} , \sqrt{5} , \sqrt{7}  etc  

Type 3 :- 

    • if m is a positive integer which is not a perfect cube then \sqrt[3]{m} is irrational . thus  \sqrt[3]{2} , \sqrt[3]{3} , \sqrt[3]{5}  etc

Type 4 :- \pi is irrational number whereas \frac{22}{7}   is rational number.

Determining the nature of the decimal expansion of rational numbers.

there are few theorem by which we will learn that when the decimal expansion of rational number is terminating and when it is non-terminating repeating. that theorem is as following.

Theorem 1 :- 

                    let x be a number whose decimal expansion terminates. then, x can expressed in the form of \frac{p}{q} , and the prime factorization of q is of the form 2^m  \times  5^n , where m , n are non-negative integers.

let’s see the converse of this theorem is true or not .if we have a rational number of the form \frac{p}{q} , and the prime factorization of q is of the form 2^m  \times  5^n , where m , n are non-negative integers. then does \frac{p}{q} have terminating decimal?

let \frac{a}{b} be rational number in the lowest form such that the prime factorization of b is the form 2^m  \times  5^n , where m , n is non-negative integers 

then we have following case  :- 

Case 1 :- when m = n then ,

\frac{a}{b}  = \frac{a}{2^m  \times  5^n}  =  \frac{a}{2^m  \times  5^m}  = \frac{a}{(10)^n}

Case 2 :- when m > n then 

    m =  n + p , where p is positive integer.

\frac{a}{b}  = \frac{a}{2^m  \times  5^n}  =  \frac{a \times  5^p}{2^m  \times  5^{m+p} }  = \frac{a \times  5^p}{2^m  \times  5^m}  =  \frac{a \times 5^p}{(2 \times 5)^n} =  \frac{c}{10^m}

Theorem 2 :- 

                  let x = \frac{p}{q}   be a rational number , such that the prime factorization of q is of the form 2^m  \times  5^n   , where m , n are non-negative integers . then x has a decimal expansion which terminates after k places of decimal , where k is the larger of m and n 

example :- (i) \frac{5}{3} = 1.666……..       (ii) \frac{17}{6}   = 2.8333………

Theorem 3 :- 

                let x = \frac{p}{q} be a rational number , such that the prime factorization of q is not of the form 2^m  \times  5^n   , where m , n are non-negative integers. then x has a decimal expansion which is non-terminating repeating.

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